What is the integral of 1/(1+3x^2) ?

The integral of 1/(1+3x^2) is 1/(sqrt(3))arctan(sqrt(3)x)+C

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integral of 1/(1+3x^2)

\int \frac{1}{1 + 3 x ^{2}} d x

\frac{1}{3} arctan (3 x) + C

Solution steps

\int \frac{1}{1 + 3 x ^{2}} d x

Apply Integral Substitution

= \int \frac{1}{3 ( 1 + u ^{2} )} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{3} \cdot \int \frac{1}{1 + u ^{2}} d u

Use the common integral:

\int \frac{1}{1 + u ^{2}} d u = arctan (u)

= \frac{1}{3} arctan (u)

Substitute back

u = 3 x

= \frac{1}{3} arctan (3 x)

Add a constant to the solution

= \frac{1}{3} arctan (3 x) + C

\frac{d}{d x} (\frac{3 x}{s q u ( x , q ) a r ( x , q ) e x ^{2} + 1})

x \to 1 lim (\frac{1}{x ^{3}})

\int sin^{3} (2 x) cos (2 x) d x

\int \frac{6}{x ^{2} + 16} d x

y^{^{''}} - 8 y^{^{'}} + 15 y = 0, y (0) = 1, y^{^{'}} (0) = 4

What is the integral of 1/(1+3x^2) ?
The integral of 1/(1+3x^2) is 1/(sqrt(3))arctan(sqrt(3)x)+C